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I have below problem:

Find $\bf C$ to minimize $\|\mathbf A-\mathbf B\mathbf C\|_F$.

Given ${\bf B} \in \mathbb R^{m \times n}$, ${\bf B}$ has lin. ind. col.

A satisfies: ${\bf DA} = {\bf E}$ , ${\bf D} \in \mathbb R^{m \times m}$, ${\bf D}$ has lin. ind. col. ${\bf E} \in \mathbb R^{m \times r}$

Thanks!

dale
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2 Answers2

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If $A=[a_1,\ldots,a_r]$ and $C=[c_1,\ldots,c_r]$ are column partitionings of $A$ and $C$, respectively, we have $$ \|A-BC\|_F^2=\sum_{i=1}^r\|a_i-Bc_i\|_2^2. $$ Hence minimizing $\|A-BC\|_F$ is equivalent to minimizing separately the Euclidean norms of $a_i-Bc_i$, $i=1,\ldots,r$. Since the least squares solution of the "single-vector" problems are given by $c_i=B^\dagger a_i$, where $B^\dagger:=(B^TB)^{-1}B^T$ is the Moore-Penrose inverse of $B$, we have $$ C=B^\dagger A. $$

Algebraic Pavel
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To expand on Pavels answer, we can use the Kronecker product and vectorization to express matrix multiplication as matrix-vector multiplication. Then we can use a suitable pseudo-inverse to find a least-squares solution.

mathreadler
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  • True but IMHO using the Kronecker product formulation is a bit an overkill as the equivalence between minimizing the Frobenius norm of a matrix expression and the sum of squares of 2-norms of its columns follows directly from the definition of the Frobenius norm and a suitable grouping of row/column sums. In addition, since the columns of $B$ are independent, there is just the least squares solution provided by the Moore-Penrose inverse. – Algebraic Pavel Apr 29 '15 at 13:33
  • It is overkill for solving the specified problem, but the objective as I saw it was to to try and get people introduced and maybe excited about the general methods so they would want to learn more. – mathreadler Apr 29 '15 at 13:44